How do I compute the matrix exponential of a nilpotent matrix?

Question

I am trying to compute the matrix exponential $e^{At}$ of the nilpotent matrix

$$ A = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$$

I have computed the eigenvalues, which are $\lambda=0$ with algebraic multiplicity $2$, and the generalized eigenvectors are

$$ u_1 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \qquad u_2 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$

How do I proceed now for computing the matrix exponential?

so, should it be simply $e^{At}=I+At$? And is it a general way of operating? Thanks. — dcr, May 27 '21 at 09:00
The general formula for matrices is $\exp(A)=I+A+\frac{1}{2}A^2+\frac{1}{6}A^3\cdots$ as you know. — Dietrich Burde, May 27 '21 at 09:01
sorry if I insist on this topic. I would like to ask if, in the case of a non-diagonalizable matrix, I could instead put it in the Jordan canonical form $J$, and so do somenting like$e^{Jt}$. The result is different, but would it be correct? Thanks. — dcr, May 30 '21 at 19:16

score 3 · Accepted Answer · answered May 27 '21 at 13:38

3

The solution for this precise $A$ is written in the above remarks.

More general, if you know that the set of all generalized eigenvectors still spans the whole space, you can still "pseudo-diagonalize" such a matrix. Hereby, by transforming into a suitable basis, you don't get rid of all off-diagonal entries, but still you can explicitly compute the exponentials. Maybe you want to look up

https://en.wikipedia.org/wiki/Jordan_normal_form

answered May 27 '21 at 13:38

nicrot000

1,507

So you mean I could do somehthing like : $e^{At}=Te^{Jt}T^{-1}$ where $J$ is the Jordan canonical form? – dcr May 30 '21 at 19:26
Precisely. Then the exponential preserves the Block structure and you can compute the exponential of a Jordan block. – nicrot000 May 31 '21 at 07:02

How do I compute the matrix exponential of a nilpotent matrix?

1 Answers1